A 1.21 kg block slides down a 25.0 m long 25.0° incline at constant velocity. How much work is done by friction?
work= force dot distance= mg*15.0*cos 25
Where is the 15 coming from? I thought that you may have meant 25 so I calculated 1.21*9.8*25*cos 25 and I got the answer wrong.
Yea, I was wondering the same thing because mine is wrong too!
To find out how much work is done by friction, we need to understand the concept of work and the forces acting on the block.
Work is defined as the product of the applied force and the displacement of the object in the direction of the force. In this case, the applied force is due to friction.
When the block is moving at a constant velocity, the net force acting on it is zero. This means that the force of friction must be equal in magnitude and opposite in direction to the component of the gravitational force acting parallel to the incline.
To calculate the force of friction, we first need to find the component of the gravitational force acting parallel to the incline. The component of the gravitational force parallel to the incline can be calculated using the following formula:
F_parallel = m * g * sin(θ)
Where:
m = mass of the block (1.21 kg)
g = acceleration due to gravity (9.8 m/s^2)
θ = angle of the incline (25.0°)
Substituting the given values:
F_parallel = 1.21 kg * 9.8 m/s^2 * sin(25.0°)
Next, we calculate the work done by friction using the formula:
Work = Force * Displacement * cos(θ)
Since the block is moving down the incline, the force of friction is acting in the opposite direction to the displacement. Therefore, we take the negative sign of the displacement:
Displacement = -25.0 m
Finally, we can calculate the work done by friction:
Work = -F_parallel * Displacement * cos(θ)
Substituting the values we obtained:
Work = - (1.21 kg * 9.8 m/s^2 * sin(25.0°)) * (-25.0 m) * cos(25.0°)